Eckart−Sayvetz conditions revisited Viktor Szalay Citation: The Journal of Chemical Physics 140, 234107 (2014); doi: 10.1063/1.4883195 View online: http://dx.doi.org/10.1063/1.4883195 View Table of Contents: http://scitation.aip.org/content/aip/journal/jcp/140/23?ver=pdfcov Published by the AIP Publishing Articles you may be interested in Rovibronically selected and resolved two-color laser photoionization and photoelectron study of titanium monoxide cation J. Chem. Phys. 138, 174309 (2013); 10.1063/1.4803161 A new ab initio intermolecular potential energy surface and predicted rotational spectra of the Kr−H2O complex J. Chem. Phys. 137, 224314 (2012); 10.1063/1.4770263 Toward an improved understanding of the AsH2 free radical: Laser spectroscopy, ab initio calculations, and normal coordinate analysis J. Chem. Phys. 137, 224307 (2012); 10.1063/1.4769778 Ab initio rovibrational structure of the lowest singlet state of O2-O2 J. Chem. Phys. 137, 114304 (2012); 10.1063/1.4752741 The rotational spectrum and dynamical structure of LiOH and LiOD: A combined laboratory and ab initio study J. Chem. Phys. 121, 11715 (2004); 10.1063/1.1814631

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THE JOURNAL OF CHEMICAL PHYSICS 140, 234107 (2014)

Eckart−Sayvetz conditions revisited Viktor Szalaya) Institute for Solid State Physics and Optics, Wigner Research Centre for Physics, Hungarian Academy of Sciences, P. O. Box 49, H-1525 Budapest, Hungary

(Received 15 April 2014; accepted 2 June 2014; published online 17 June 2014) It is shown that vibrational displacements satisfying the Eckart−Sayvetz conditions can be constructed by projection of unconstrained displacements. This result has a number of interesting direct and indirect ramifications: (i) The normal coordinates corresponding to an electronic state or an isotopologue of a molecule are transformed to those of another state or isotopologue by a linear and, in general, non-orthogonal transformation. (ii) Novel interpretation of axis switching. (iii) One may enhance the separation of rotational-large-amplitude internal motions and the vibrational motions beyond that offered by the standard use of the Eckart−Sayvetz conditions. (iv) The rotationalvibrational Hamiltonian given in terms of curvilinear internal coordinates may be derived with elementary mathematical tools while taking into account the Eckart conditions with or without enhancement. © 2014 AIP Publishing LLC. [http://dx.doi.org/10.1063/1.4883195] I. INTRODUCTION 1–3

It is beautifully described in Herzberg’s classic books how the idea of separating the motions of a molecule as electronic, rotational, and vibrational motions emerges from the analysis of spectra. It is also pointed out there that the model consisting of a rigid rotor and harmonic oscillators is an approximation, and more accurate analyses and theoretical prediction of rotational-vibrational spectra should take into account, besides anharmonic effects, interactions of the rotational and vibrational motions. To grasp how rotation and vibrations and their interaction arise, one has to follow the development of the nuclear motion Hamiltonian, classical and/or quantum. The vibrations of many molecules are of small amplitude and take place around the equilibrium structure of the molecules. These molecules are called semirigid molecules. Often, there are one, two, or more vibrational modes having large-amplitude. Large-amplitude internal motions (LAMs) may connect two or more minima on the potential energy surface. Molecules with LAMs are called non-rigid molecules.4 The description of the nuclear motions of a semirigid molecule as overall rotation, vibrations, and their interactions is facilitated by the Eckart conditions. RotationalLAM and vibrational motions may be delineated by the use of the Eckart−Sayvetz (ES) conditions5, 6 in the case of nonrigid molecules. The Eckart and the ES conditions play a fundamental role in developing the rotational-vibrational and the rotational-LAM-vibrational Hamiltonian of semirigid and non-rigid molecules, respectively.4, 7–9 These conditions ensure separation of the rotational and the vibrational motions in the case of semirigid molecules and the separation of the rotational-LAM and the vibrational motions in the case of non-rigid molecules, in the zero order of approximation, that is, when all small amplitude vibrational displacements are zero. The ES conditions along with some of their consea) Email: [email protected]

0021-9606/2014/140(23)/234107/12/$30.00

quences are the central theme of this communication. A more general discussion about the separation of rotations and vibrations can be found in Ref. 10. A review of the ES conditions will be given in Sec. II. It is, however, not a recollection of past contributions, but an attempt to give a self-contained explanation of how the rotational-LAM and the vibrational motions can be separated. As a whole our discussion of the ES conditions gives new point of view with placing emphasize on constructing displacements satisfying the ES conditions (ES displacements) via projection. This construction leads to new results related to a number of practical questions. In particular, Sec. III considers how normal coordinates corresponding to two different electronic states or two isotopologues of a molecule can be transformed into each other. Section IV discusses the related results of earlier approaches. Section V gives a new look at the axis switching transformation. It is shown in Sec. VI that one can improve the separation of the rotational-LAM and the vibrational motions beyond that one can achieve by the standard use of the ES conditions. Section VII sketches the steps of developing a rotational-vibrational Hamiltonian in terms of curvilinear internal coordinates while satisfying the Eckart conditions. A number of authors have addressed this question in the case of molecules with three11–16 and more atoms.17–22 The advantages of the derivation proposed here are its simplicity in imposing the Eckart conditions and the possibility for improving rotational-vibrational separation beyond that achievable by the standard use of the Eckart conditions. The results are summarized in Sec. VIII.

II. CALCULATION OF VIBRATIONAL DISPLACEMENTS SATISFYING THE ES CONDITIONS

Initially, we shall use displacement coordinates, dαn for describing small amplitude vibrations. We shall think of dαn as variables which take well-defined values at any instant of time determined by a set of equations, Eq. (5),

140, 234107-1

© 2014 AIP Publishing LLC

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234107-2

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

including the ES conditions, Eqs. (5b)–(5d). Since general displacements do not satisfy the ES conditions, this set of equations may be thought of as the defining equations of ES displacements. Though correct, this definition is too complicated, as explained in Sec. II A. In Sec. II B, it will be shown ES , can be derived by simple projecthat ES displacements, dαn tion of general (unconstrained) displacements. A. Notions, notation, and a complicated method for determining ES displacements

Consider a nonlinear molecule of N atoms. The motion of the molecule may be described by employing the instantaneous position vectors of its constituent particles, its electrons and nuclei. Let us introduce a laboratory system (LS) of axes, that is, a Cartesian coordinate system fixed in space. Let rαk and uαn denote the LS coordinates, α = X, Y, Z, of the instantaneous position vector of the kth electron and the nth nucleus, respectively. In principle, one could use these coordinates to describe the motion of the molecule. Motivated by practical and physical reasons, however, one distinguishes overall translation, overall rotation, and internal motions of the molecule. One also attempts to minimize coupling between the motions of various type. To describe overall translation, the LS coordinates of the center of mass of the nuclei, Rα , can be used. Overall rotation is described as change of orientation of a Cartesian system of axes attached to the molecule (molecule fixed system of axes (MS)) with respect to the LS. It can be parametrized by three Euler angles, θ , φ, χ . The Euler angles serve as rotational coordinates. Thus, for the coordinates of electrons we have the relationships rαk = Rα + [S−1 (θ, φ, χ )]αβ eβk ,

(1)

where eβk denotes the MS coordinates (β = x, y, z) of the instantaneous position of the kth electron, and summation over repeated Greek indices has been assumed and will be assumed throughout the paper. Motion of the electrons is approximately decoupled from that of the nuclei by the BornOppenheimer (BO) approximation. In what follows validity of the BO approximation is assumed and attention is paid to the nuclear motions. For the coordinates of the nuclei equations similar to that of Eq. (1) of the electrons, uαn = Rα + [S−1 (θ, φ, χ )]αβ rβn ,

mn aβn = 0

N 

αβγ

N 

n=1

(4)

required, where mn denotes the mass of the nth nucleus. A nonlinear N-atom molecule possesses 3N − 6 internal degrees of freedom. Accordingly, its internal motions may

(5b)

mn aβn dγ n = 0,

(5c)

∂aαn dαn = 0, ∂ρi

(5d)

n=1 N 

n=1

mn dβn = 0,

n=1

(3)

with N 

uαn = Rα + [S−1 (θ, φ, χ )]αβ [aβn (ρ1 , ρ2 , . . . , ρL ) + dβn ], (5a)

(2)

hold, where rβn stands for the MS coordinates of the instantaneous position of the nth nucleus. Then, rβn are written as the sum of the coordinates of the nuclei of a reference configuration, aβn , and displacements coordinates dβn , rβn = aβn + dβn ,

be described by 3N − 6 curvilinear internal coordinates ρ i , i = 1, 2, . . . , 3N − 6. For a semirigid molecule, a rigid reference configuration is chosen. It corresponds to a configuration of the molecule when the internal coordinates are fixed at constant values, usually at their equilibrium values ρ 0i . Thus, the MS coordinates of the rigid reference con0 , are of constant values. For non-rigid figuration, aβn = aβn molecules, that is, for molecules with one or more, say L, LAMs, a non-rigid reference configuration is chosen. The structure of the non-rigid reference configuration may be defined by its internal coordinates either as (a)ρ1 ≡ ρ 1 , ρ2 ≡ ρ 2 , . . . , ρL ≡ ρ L , ρ 0L+1 , ρ 0L+2 , . . . , ρ 03N−6 , or as (b) ρ1 (= ρ 1 ), ρ2 (= ρ 2 ), . . . , ρL (= ρ L ), ρ 0L+1 , ρ 0L+2 , . . . , ρ 03N−6 . (c) One might also employ least squares LAM coordinates.24 In the case of definition b, the internal coordinate ρ i of the reference configuration is equivalent to the internal coordinate ρ i of the molecule, but they are not identical, since their instantaneous values may differ. Then, one speaks of effective LAM coordinates. LAM coordinates of type a are called real LAM coordinates.9 The non-rigid reference configuration owes its name to the fact that its MS coordinates, aβn = aβn (ρ 1 , ρ 2 , . . . , ρ L ), depend on the L large-amplitude internal coordinates. Thus, we have defined 3N vibrational coordinates (dβn ) for an N-atom semirigid molecule and 3N + L vibrational and LAM coordinates (dβn , ρ i ) for an N-atom non-rigid molecule. Since there are only 3N − 6 internal degrees of freedom, one must introduce 6 and 6 + L additional equations for semirigid and non-rigid molecules, respectively, to constrain the displacements dβn . One may choose the constraint equations by requiring minimal coupling between vibrational and overall motions and between the vibrational and large-amplitude internal motions. These constraints are called the Eckart conditions, Eqs. (5b) and (5c), and Sayvetz conditions, Eq. (5d), respectively. The same constraint equations complete the 3N equations relating the instantaneous LS coordinates of nuclei to the 3N + 6 + L coordinates of overall, vibrational, and large-amplitude internal motions. That is, given the coordinates uαn one can calculate the values of the translational, rotational, vibrational, and LAM coordinates Rα , θ, φ, χ , dβn , ρ1 , ρ2 , . . . , ρL by solving the system of equations:

mn

i = 1, 2, . . . , L,

(5e)

where  denotes the Levi-Civita tensor. Note that the equations related to a semirigid molecule are special cases of those corresponding to a non-rigid molecule. Therefore, it

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234107-3

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J. Chem. Phys. 140, 234107 (2014)

suffices to discuss only the case of non-rigid molecules in detail. When LAM coordinates of type a are employed the Sayvetz conditions are obeyed automatically.9 Here, we shall assume effective LAM coordinates (LAM coordinates of type b). Then satisfaction of the Sayvetz conditions must be enforced. When solving a system of equations one can multiply, add, and subtract the equations. With suitable multiplications, additions, and subtractions of Eqs. (5a), and by using Eq. (4), one can obtain the system of equations: N 1  mn uαn , Rα = M n=1

αβγ

N 

mn aβn Sγ α (uαn − Rα ) = 0,

ing to solve the equations of Eckart axis conditions and Eqs. (6a)–(6c). Let us introduce 3 non-genuine translational, Tα , 3 nongenuine rotational, Rα , L non-genuine LAM, Pi , and 3N − 6 − L vibrational, Qr , coordinates by the equations: Tα =

N 

Rα = αβγ

N 

Pi =

N 

mn

n=1

n=1 N 

Qr = γr−1/2 qr =

∂aβn mn [Sβα (uαn − Rα ) − aβn ] = 0, ∂ρi n=1

(6c)

dβn = Sβα (uαn − Rα ) − aβn ,

(6d)

(7a)

mn aβn dγ n ,

(7b)

∂aαn dαn , ∂ρi

(7c)

n=1

(6a)

(6b)

mn dαn ,

n=1

N 

m1/2 n lαn,r dαn ,

(7d)

n=1

where dβn are general, unconstrained, displacements,

 where M = N n=1 mn . The system of equations (6) is equivalent to Eqs. (5). Equation (6a) gives the coordinates of translation. By using them one can solve Eq. (6b) to obtain the coordinates, as functions of ρ i , of overall rotation. One proceeds by solving Eq. (6c) to obtain the instantaneous values of the LAM coordinates. Then, by substituting these values into the LAM dependent functions determined previously, one obtains the instantaneous values of the rotational coordinates. When the instantaneous values of translational, rotational, and LAM coordinates have been calculated one can calculate the instantaneous values of the vibrational displacements dβn satisfying the ES conditions. The calculation of Rα is obvious. For the calculation of Sγ α , that is for solving the Eckart axis conditions, Eq. (6b), useful methods have been worked out.25–32 However, Eq. (6c) can be rather difficult to solve. Once the instantaneous values of translational, rotational, and LAM coordinates have been obtained, the calculation of vibrational displacements obeying the ES conditions is simple, but overall this is a rather complicated procedure. In fact, note that four equations, Eq. (6d) along with Eqs. (6a)–(6c), define the ES displacements. A significantly simpler method than the procedure described above to define ES displacements will be introduced in Subsection II B. In particular, it will be shown that ES displacements can be defined in terms of new vibrational coordinates, Qr , without referring to the Eckart axis conditions and Eqs. (6a)–(6c). In passing it is worth pointing out that besides giving the instantaneous values of the translational, rotational, vibrational, and LAM coordinates, another important application of Eq. (6) is in determining how these coordinates transform under the permutation-inversion group of the molecule.8

2π cωr (8) ¯ with c denoting the speed of light in vacuum, ¯ denoting Planck’s constant, h, divided by 2π , and ωr is a quantity of dimension of wavenumbers, and lαn, r are linear combination coefficients to be determined later. Note, that with γ r so defined qr is dimensionless. We shall invert Eq. (7). The set of mass-weighted Cartesian displacements 1/2 mn dαn can be thought of as a set of orthonormal basis vectors spanning a 3N-dimensional vector space. Then, Tα , Rα , Pi , and Qr may be considered as vectors in this vector space. They can be represented as column vectors of their coordinates obtained in the basis of mass-weighted Cartesian displacements. These coordinates can be read from Eq. (7), γr =

Tα,βn = m1/2 n δαβ ,

(9a)

Rα,βn = αγβ m1/2 n aγ n ,

(9b)

Pi,βn = m1/2 n

∂aβn , ∂ρi

Qr,βn = lβn,r .

(9c) (9d)

By direct calculations using these coordinates one obtains that Tα TβT = Mδαβ , Tβ RTα = Rα TβT = 0, Tα PiT = Pi TαT = 0,

(10a) (10b)

Rα RTβ = Iαβ ,

(10c)

B. A simple method for calculating ES displacements

Rα PiT = Iαi , Pi RTα = Iiα ,

(10d)

Below it will be shown how vibrational displacements obeying the ES conditions can be obtained without hav-

Pi PjT = Iij ,

(10e)

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234107-4

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

where superscript T denotes transposition and Iαβ , Iαi , Iiα , and Iij are elements of the 3 + L dimensional generalized tensor of inertia I defined as Iαβ = αγ  βδ

N 

mn aγ n aδn ,

(11a)

n=1

Iαi = Iiα = εαβγ

N 

mn aβn

∂aγ n , ∂ρi

N 

∂aαn ∂aαn mn . Iij = ∂ρi ∂ρj n=1

(11b)

LT¯α ,γ n = T¯α,γ n = M −1/2 m1/2 n δαγ ,

[I−1/2 ]αi m1/2 n

i=1

[I−1/2 ]ij m1/2 n

j =1

∂aγ n , ∂ρi

∂aγ n , ∂ρj

LQr ,γ n = Qr,γ n = lγ n,r .

(12a)

m1/2 n lαn,r = 0,

(12b)

(12c)

(13a)

m1/2 n aβn lγ n,r = 0,

(13b)

∂aαn lαn,r = 0, ∂ρi

(13c)

m1/2 n

i = 1, 2, . . . , L, N  n=1

where Pαn,βp = δnp δαβ −



LT¯γ ,αn LT¯γ ,βp

γ =x,y,z

−



LR¯ γ ,αn LR¯ γ ,βp −

γ =x,y,z

L 

LP¯ i ,αn LP¯ i ,βp . (16)

i=1

¯ α , P¯ i } Since by way of their construction the vectors {T¯α , R T ¯ T ¯ T ¯ ¯ ¯ ¯ are orthonormal, the matrices Tα Tα , Rα Rα , and Pi Pi are (orthogonal) projection matrices and so is the matrix L 

P¯ iT P¯ i ,

(17)

whose detailed expression is just Eq. (16). Thus, we have shown that in order for L to be orthonormal the linear combination coefficients lαn, r must satisfy the sum rules expressed by Eqs. (13) and (15). Shortly, it will be described how such lαn, r can be calculated. Note that the projection matrix, P, given in Eq. (15) is completely determined by the atomic masses and the coordinates of the reference configuration. Due to Eq. (4) P can be factorized such that

(12d)

n=1

n=1

(15)

P = PT¯ PR¯ P¯ 1 ...P¯ L = PR¯ P¯ 1 ...P¯ L PT¯ ,

(18)

where PT¯ and PR¯ P¯ 1 ...P¯ L project onto the subspace comple¯ P¯ 1 , . . . , P¯ L , respecmentary to that spanned by T¯ and R, tively. Furthermore, lαn,r =

N 

Pαn,βp lβp,r .

(19)

p=1

N 

N 

(14)

i=1

n=1

εαβγ

QTr Qr

r=1

¯TR ¯ P = 1 − T¯αT T¯α − R α α −

¯ α , P¯ i , Qr } is an orthonormal set if The set of vectors {T¯α , R T T ¯ ¯ Tα Qr = 0, Rα Qr = 0, P¯ i QTr = 0, and Qr QTs = δrs . These equations are satisfied if the linear combination coefficients lαn, r satisfy the sum rules N 

3N−6−L 

r=1

LP¯ i ,γ n = P¯ i,γ n = [I−1/2 ]iδ εδβγ m1/2 n aβn L 

P¯ iT P¯ i +

lαn,r lβp,r = Pαn,βp ,

(11c)

¯ α,γ n = [I−1/2 ]αδ εδβγ m1/2 aβn LR¯ α ,γ n = R n

+

L 

should also be required. Arranged in a more informative form and written out in detail this requirement reads as 3N−6−L 

One can see that the vectors Tα , Rα , Pi , Qr are not orthonormal. Orthonormal vectors are, however, preferred, since they can be related to the mass-weighted Cartesian displacements by an orthogonal transformation which is easy to invert. Orthonormalization of Tα , Rα , Pi gives an orthonormal ¯ α , P¯ i . set of vectors of non-genuine internal coordinates T¯α , R Then one can find that the elements of the matrix L transforming the mass-weighted Cartesian displacements to the ¯ α , P¯ i , Qr }, are given by the new basis vectors, the set {T¯α , R expressions:

L 

¯TR ¯ 1 = T¯αT T¯α + R α α +

i=1

n=1

+

If the sum rules given above are satisfied, LLT = 1, where 1 denotes the 3N by 3N unit matrix. In order for L to be orthonormal LT L = 1, that is,

lαn,r lαn,s = δrs .

(13d)

The projection matrix P has been derived and employed to specific problems in Refs. 9, 33, and 34. Since L is orthonormal, one may express the massweighted, unconstrained Cartesian displacements in terms of ¯ α , P¯ i , Qr as the coordinates T¯α , R   ¯β m1/2 LT¯β ,αn T¯β + LR¯ β ,αn R n dαn = β=x,y,z

β=x,y,z

L 

3N−6−L 

+

i=1

LP¯ i ,αn P¯ i +

LQr ,αn Qr .

(20)

r=1

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234107-5

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

Now set the values of the non-genuine internal coordinates to zero and substitute the constrained displacements ES m1/2 n dαn =

3N−6−L 

LQr ,αn Qr =

3N−6−L 

r=1

lαn,r Qr

(21)

r=1

so obtained into Eq. (5). Observe, that due to Eqs. (13a)–(13c) ES the displacements dαn obey the ES conditions, which is emphasized by the superscript ES. By substituting Eq. (7d) into Eq. (21), one obtains ES m1/2 n dαn =

N 

r Impose the ES conditions. Imposing the ES conditions amounts to substituting the unrestricted massweighted Cartesian displacements with mass-weighted ES displacements. r Build the invariance property expressed by Eq. (23) into the equations. The approximate potential energy surface so obtained, Vappr , is

Pαn,βp m1/2 p dβp

(22)

ES to unconstrained displacerelating the ES displacements dαn ments. Since PP = P, we also have ES m1/2 n dαn =

ES Pαn,βp m1/2 p dβn .

+

V (ρ¯1 , ρ¯2 , . . . , ρ¯3N−6 ) ≈ V0 +

N 

N 

m−1/2 Fαn Pαn,βp , n

(27)

n=1

+

Hγp,δs =

N 

Pαn,γp m−1/2 Fαn,βo m−1/2 Pβo,δs . (28) n o

n,o=1

r Then, use Eq. (21). The result is Vappr = V0 +

3N−6−L  r=1

1 + 2

3N−6−L 

= V0 +

⎣

⎡ ⎣

r,t=1 3N−6−L  r=1

⎡

N 

⎤

Fβp lβp,r ⎦ Qr

p=1 N 

⎤ lγp,r Hγp,δs lδs,t ⎦ Qr Qt

s,p=1

r Q r +

1 2

3N−6−L 

rt Qr Qt .

r,t=1

(29) These equations also serve as the defining equations for the coefficients r and rt .

Fαn dαn

n=1 N 1  Fαn,βo dαn dβo . (24) 2 n,o=1

Since the reference configuration is non-rigid, V0 , Fαn , and Fαn, βo are functions of the LAM coordinates ρ i . Let us transform the approximate potential energy surface, Eq. (24), to the vibrational coordinates Qr . The transformation is carried out in a number of steps:

r Transform to mass-weighted Cartesian displacement coordinates V (ρ¯1 , ρ¯2 , . . . , ρ¯3N−6 )   m−1/2 Fαn m1/2 n n dαn

n=1

+

Fβp = and

It should be clear that the role of the ES conditions is to restrict the allowable Cartesian displacements to those given in Eq. (21). Displacements satisfying the ES conditions can be obtained according to Eq. (22) by projection of unconstrained displacements by a projection matrix determined entirely by the reference configuration and the atomic masses. This is a much simpler way of obtaining ES displacements than the approach described in Sec. II A and it allows one to build the ES conditions in a Hamiltonian (Lagrangian) without explicitly calculating the values of the ES displacements. To complete the derivations, we have yet to find a way of calculating the linear combination coefficients lαn, r satisfying Eqs. (13) and (15). This will be done next. Expand the potential energy surface of the molecule with respect to unrestricted Cartesian displacements around the non-rigid reference configuration up to the quadratic terms

N 

N   1/2 ES  1  ES Hγp,δs m1/2 p dγp ms dδs , (26) 2 s,p=1

where

(23)

p=1

≈ V0 +

  ES Fβp m1/2 p dβp

p=1

p=1

N 

N 

Vappr = V0 +

N 

 1/2  1/2  1 mn dαn mo dβo . m−1/2 Fαn,βo m−1/2 n o 2 n,o=1 (25)

The projection matrix P is a representation of a projection operator Pˆ which projects from the 3N-dimensional ¯ α , P¯ i , Qr } onto the 3N space spanned by the vectors {T¯α , R − 6 − L dimensional subspace spanned by the Qr s. In addi1/2 1/2 ES (Eq. (22)). According tion, P projects mn dαn onto mn dαn to Eq. (21) the latter are in the space spanned by the Qr s. Since the ES displacements are in the Qr space and satisfy the ES conditions, the sum rules given in Eqs. (13a)–(13c) are obeyed. In other words, Eqs. (13a)–(13c) are satisfied once the ES conditions have been imposed. Imposing the ES conditions was in fact the second step in the derivation of Eq. (29). The linear combination coefficients also have to obey the sum rules given in Eqs. (13d) and (15). In the third step in deriving Eq. (29) we introduced, by employing Eq. (23), the projected, mass-weighted quadratic force matrix H. This matrix has 6 + L zero and 3N − 6 − L non-zero eigenvalues. Observe that the eigenvectors of H are orthonormal and the projection matrix formed by the eigenvectors corresponding to non-zero eigenvalues is given in the same basis and projects onto the same 3N − 6 − L dimensional

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234107-6

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

subspace as the projection matrix P. Therefore, lαn, r defined as elements of the eigenvectors corresponding to the nonzero eigenvalues of H satisfy, in addition to Eqs. (13a)–(13c), Eqs. (13d) and (15). Thus, we have shown how suitable lαn, r can be calculated. In addition, it follows that with the lαn, r so calculated the matrix of elements rt is diagonal, that is, with λ2r denoting the nonvanishing eigenvalues of H, rt = λ2r δrt ,

(30)

where δ rt is the Kronecker delta symbol. Thus, the potential energy surface may be approximated as Vappr = V0 +

3N−6−L  r=1

1 r Q r + 2

3N−6−L 

λ2r Q2r .

(31)

r=1

It can be rewritten in terms of the dimensionless coordinates qr , Vappr = V0 +

3N−6−L 

φr qr +

r=1

with coefficients

φ r = r

1 2

¯ 2π cωr

φrr = λ2r

3N−6−L 

φrr qr2 ,

(32)

interesting question is how to transform between normal coordinates of the isotopologues of a molecule.35 These questions can be answered on the same footing. Let A and B denote either the initial and the final state of an electronic transition, respectively, or two different isotopologues of a molecule. We shall use superscript B to indicate if a quantity belongs to B. For simplicity, no superscript A is attached to quantities belonging to A. Accordingly, Eq. (29) is the potential energy corresponding to A expressed in normal coordinates of A. The relationships derived in Sec. II apply to B without change except that superscript B should be added to quantities specific to B. We shall make use of these relationships. Note, that the unconstrained displacements dαn are specific neither to A nor to B. Below it is shown that the normal coordinates of A can be related to those of B by linear transformation. We shall describe two different methods for obtaining the transformation. A. Method I

The ES displacements of A given in Eq. (21) are not ES displacements for B. Rearrange Eq. (21) as

r=1

1/2

¯ , 2π cωr

,

 B 1/2 ES dαn = mn

(33)

(34)

obtained by using the relationship (7d) connecting Qr and the dimensionless coordinate qr and the defining equation, Eq. (8), of γ r . However, ωr is yet to be defined. By requiring φ rr = ωr and employing Eq. (34) one finds that 

¯ 1/2 λr , (35) ωr = 2π c where only roots λr > 0 of λ2r are accepted as physically relevant. Clearly, Qr and qr are the familiar normal and dimensionless normal coordinates, respectively, of vibrations of small amplitude. Except for Eq. (22) each of the expressions given here has already been known. However, they are scattered through a number of papers. We have joined them in a single line of thoughts leading to a hopefully useful account of the ES conditions and to new theoretical results, to be discussed in Secs. III–VII, of practical value. The use of Eq. (22) will be decisive in the derivations to come.



mBn mn

1/2 3N−6−L 

lαn,r Qr .

(36)

r=1

Since ES displacements for B can be obtained from any displacement by projection similar to that shown in Eq. (22), we have N  B 1/2 ES  B 1/2 ES,B  B dαn = Pαn,βp dβp . mn mn

(37)

p=1

Thus, by substituting Eq. (36) into Eq. (37),

1/2 3N−6−L N  mBp  B 1/2 ES,B  B mn dαn = Pαn,βp lβp,r Qr (38) mp p=1 r=1 follows. Since, similar to Eq. (21) of A, we have   B 1/2 ES,B 3N−6−L B mn dαn = lαn,r QBr B

(39)

r=1

for B, the equality B 3N−6−L 

B lαn,r QBr

=

r=1



N 

B Pαn,βp

mBp

1/2 3N−6−L 

mp

p=1

lβp,r Qr

r=1

(40) III. RELATION OF NORMAL COORDINATES CORRESPONDING TO TWO DIFFERENT ELECTRONIC STATES OR TWO DIFFERENT ISOTOPOLOGUES OF A MOLECULE

The knowledge of the transformation between the sets of normal coordinates corresponding to two different electronic states of a molecule is of importance when analysing the vibrational structure in an electronic transition. The literature is divided as to the nature of this transformation, that is, if it is linear or nonlinear, orthogonal or non-orthogonal. Another

holds. Finally, Eq. (40) can be simplified as QBt =

3N−6−L 

ZBA tr Qr ,

(41)

r=1

where ZBA tr =

N  p=1

B lβp,t

mBp mp

1/2 lβp,r .

(42)

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234107-7

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

r Then, use Eq. (21) corresponding to B to obtain

A similar derivation shows that Qt =

B 3N−6−L 

B ZAB tr Qr ,

(43)

Vappr = V0 +

r=1

=

N 

lβp,t

p=1

  N

mp mBp

 −1/2 Fβp mBp /mp

p, ,p=1

r=1

 B B B × Pβp,γp , lγp , ,r Qr

where ZAB tr

B 3N−6−L 

1/2 B lβp,r .

(44) +

B B lβp , ,r Pγp,βp ,

s , ,p, =1

r,t=1

⎤ N   B −1/2  B −1/2 ⎦ mp /mp ×⎣ Hγp,δs ms /ms

Note that ZBA depends on the LAM coordinates of A and B, B that is, ZBA = ZBA ({ρiB }Li=1 , {ρi }Li=1 ), since the linear combination coefficients lβp, r depend on the LAM coordinates of B depend on the A and the linear combination coefficients lβp,r LAM coordinates of B. For the same reason, ZAB is a function of the LAM coordinates of A and B.

s,p=1

 B B B B × Pδs,νs , lνs , ,t Qr Qt = V0 +

B. Method II

  N

B 3N−6−L 

1 2 ⎡

B 3N−6−L 

B AB r Qr +

r=1

1 2

B 3N−6−L 

B B AB rt Qr Qt .

r,t=1

Equation (31) is an approximation to the potential energy function of A given in terms of the normal coordinates of A. Now let us transform this expression to one given in terms of the normal coordinates of B. By starting from Eq. (26) the derivation proceeds as follows:

These equations also serve as the defining equations for the coefficients AB and AB r rt .

r Note, that the displacements d ES are not ES displaceβp

Let us carry out the calculations described above in the backward direction by starting from Eq. (47):

ES ES ments for B. Transform from dβp to (mBp )1/2 dβp , i.e., to displacements weighted according to B,

Vappr = V0 +

N 

 −1/2  B 1/2 ES  mp Fβp mBp /mp dβp

p=1

+

N −1/2  −1/2 1  B mp /mp Hγp,δs mBs /ms 2 s,p=1

 1/2 ES  B 1/2 ES  × mBp dβp ms dδs .

ES,B ES by (mBp )1/2 dβp . amounts to replacing (mBp )1/2 dβp r Build in the invariance property expressed by Eq. (23) corresponding to B,

Vappr = V0 +

r Use the equation analogous to Eq. (21) to transform from normal coordinates of B to ES displacements of B. The transformation gives Eq. (46). r Use the equation analogous to Eq. (23) to eliminate the projection matrix P B , Vappr = V0 +

 −1/2 Fβp mBp /mp

+

r Use the defining equations of F and H, that is, Eqs. (27) and (28), respectively, along with Eq. (22) to obtain Vappr = V0 +

 B 1/2 ES,B  mp , dγp,

+

× mp ,

ES,B  B 1/2 ES,B  ms , dβp dνs , . ,

 ES,2  Fβp m1/2 p dβp

N   1/2 ES,2  1  ES,2 ms dδs , (49) Hγp,δs m1/2 p dγp 2 s,p=1

where ES,2 m1/2 n dαn =

  −1/2 B Pδs,νs , × Hγp,δs mBs /ms  B 1/2

N  p=1

 N N  B −1/2 1  B mp /mp Pγp,βp, 2 s , ,p, =1 s,p=1



 ES,B  Fβp m1/2 p dβp

N   1/2 ES,B  1  ES,B + ms dδs . (48) Hγp,δs m1/2 p dγp 2 s,p=1

p, ,p=1 B × Pβp,γp ,

N  p=1

(45)

r Impose the ES conditions corresponding to B. It

N 

(47)

N 

ES,B Pαn,βp m1/2 p dβp

p=1

(46)

=

N  (PP B P)αn,βp m1/2 p dβp .

(50)

p=1

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234107-8

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

r Due to the definition of lαn, r , Hγp,δs =

3N−6−L 

lγp,r λ2r lδs,r .

(51)

the relationships      LB   V {ρi }Li=1 , Q = V {ρi }Li=1 , ZAB {ρi }Li=1 , ρiB i=1 Q B , (57)

r=1

Therefore, by using Eq. (51), Eq. (49) can be rewritten as Vappr = V0 +

3N−6−L 

r + r Q

r=1

1 2

3N−6−L 

2r , (52) λ2r Q

r

where r = Q

N 

ES,2 m1/2 n lαn,r dαn .

(53)

n=1

By comparing Eq. (52) with (31) one finds that r = Qr . Q Since no approximation was introduced during the backward derivations, we have, in fact, shown that Vappr = V0 +

3N−6−L 

B

B AB r Qr +

r=1

= V0 +

3N−6−L  r=1

1 r Q r + 2

1 2

3N−6−L 

B lδs,t =

B B AB rt Qr Qt

r,t=1

3N−6−L 

λ2r Q2r .

(55)

By substituting Eqs. (51) and (55) into Eq. (47) and by using the defining equation of F one obtains Vappr = V0 +

r ,

r , =1

×

B  3N−6−L N  

r=1

+

×

3N−6−L 1  2 λ, 2 r , =1 r

t=1

×

   −1/2 B lβp,r , mBp /mp lγp, ,r QBr

p=1

B  3N−6−L N  

r=1

   −1/2 B lδs,r , mBs /ms lδs,t QBt

In previous works, semirigid molecules were considered and, by following the intuitive arguments due to Duschinsky,36 a transformation of the form 3N−6 

Lrt Qt + cr ,

(59)

where cr are constants and Lrt is the rtth element of the transformation matrix L, was looked for. Different results have been obtained concerning the properties, e.g., linear or nonlinear, of the transformation matrix L. In particular, the authors of Ref. 37 obtained a linear transformation. They built in their formulas the difference of origins of the two sets of displacements which led to the shift terms, cr , in the transformation. Normal coordinates are, however, displacement coordinates. They are to be considered as variables taking any physically allowed values. These values do not depend on where the origin of a displacement is located. Therefore, the transformation between two sets of such coordinates should not include any shift of the coordinates. This manifests in the results of Sec. III. Since, in general, electronic transitions take place instantaneously with respect to nuclear motions, one may write uαn = uBαn

s=1

B  3N−6−L N  

IV. ON EARLIER RESULTS OF NORMAL COORDINATE TRANSFORMATION

t=1

s , =1

3N−6−L 

(58)

hold, where Q and Q B denote the collection (column vector) of normal coordinates of A and B, respectively. For practical calculations, one should have a single set of LAM coordiB nates, i.e., {ρi }Li=1 ≡ {ρiB }Li=1 . Such a set may be obtained as a union of the LAM coordinates of A and B. Clearly, the transformation between the sets of normal coordinates of A and B is linear and, in general, nonorthogonal. It does not include any shift of the coordinates.

QBr =

B lαn,r ,

B B Pδs,νs , lνs , ,t .

 LB  V B ρiB i=1 , Q B  LB  LB   = V B ρiB i=1 , ZBA ρiB i=1 , {ρi }Li=1 Q

(54)

r=1

Recall that due to the definition of N 

B

and

   B −1/2 B mp /mp lγp,r lγp,r , QBr ,

p=1

(56) which exhibits the same relation, Eq. (43), as obtained in Sec. III A for connecting the normal coordinates of B to those of A. When we start with the potential of B a similar derivation leads to Eq. (41) of Sec. III A. Examination of the transformation of higher order potential energy terms leads to the same transformation formulas as obtained in Sec. III A. Therefore,

(60)

at the moment of the transition. However, there is no physical justification for using Eq. (60) when A and B are isotopologues, for their dynamics are different and independent of each other. Since a molecule in an electronic state A may follow dynamics different from and independent of that in its electronic state B, there is no physical justification for assuming Eq. (60) neither before nor after the transition. Clearly, Eq. (60) is a valid assumption only at the moment of transition. A consequence of Eq. (60) is that the MS coordinates of A and B are related by rotation T −1 = S−1 S,  B ES,B  ES aβn + dβn . (61) = (T −1 )βδ aδn + dδn

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234107-9

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

Being valid only at a given instant of time these equations can be used to establish a relationship, valid only at this particular instant of time, between the normal coordinates corresponding to electronic states A and B. However, Eq. (61) cannot establish a relationship which is valid at any instant of time. It is exactly this questionable use of Eq. (61) which has led to nonlinear transformation between the two sets of normal coordinates.38, 39 V. AXIS SWITCHING

By taking the cross product of Eq. (61) with respect to mn an (an = (axn , ayn , azn )T ), followed by summation with respect to n leads to the equation40 αβγ

N 

 B ES,B  = 0, mn aβn (T −1 )γ δ aδn + dδn

(62)

n=1

whose solution gives the sudden rotation, the switching, from the MS corresponding to the electronic state A to the MS corresponding to state B at the moment of transition from A to B. It is important to emphasize that T is a rotation with fixed values of the Euler angles. The rotation is sudden and takes place instantaneously on the time scale of nuclear motions. AccordES,B , nor ingly, it can be thought of as a function of neither dδn B as a function of the related normal coordinates Q , nor as a function of the LAMs. How can one solve Eq. (62) without knowing the actual values of the vibrational and LAM coordinates at the moment of transition? This puzzle may be solved as follows. Consider the structure of the molecule at the moment of the electronic transition as a reference configuration for the final state (B), from which, after the transition, the vibrations and LAMs set on. As long as no approximation is made in describing the dynamics, the results of calculations should be independent of this configuration. Therefore, we are allowed to assume zero values for the vibrational coordinates and to fix the LAM coordinates at their equilibrium values in Eq. (62). Then Eq. (62) can be solved to obtain the axis switching transformation. Sometimes the transformation so derived is called zero order axis switching.39 We have pointed out why this is the physically correct and complete transformation. VI. ENHANCED ES SEPARATION

Equation (22) can be written in a slightly more compact form ES,1 m1/2 = n dn

N 

recurrence relation ES,j +1 = m1/2 n dn

N 

ES,j Vn,p m1/2 p dp .

(65)

p=1

Since P and A do not commute, the vectors ⎛ 1/2 ES,j ⎞ m1 d 1 ⎜ 1/2 ES,j ⎟ ⎜m d ⎟ ⎜ 2 2 ⎟ ⎜ ⎟ , j = 1, 2, . . . ⎜ ⎟ .. ⎜ ⎟ . ⎝ ⎠

(66)

1/2 ES,j

mN d N

are different vectors. Since, by way of construction, they obey 6 + L ES conditions, only 3N − 6 − L of them are linearly independent. Accordingly, the general form of ES displacements is d ES n =

3N−6−L 

cj d ES,j , n

(67)

j =1

where cj are linear combination coefficients. Since d ES n satisfy the ES conditions several terms coupling the translational, rotational-LAM, and vibrational motions can be removed from the Hamiltonian in the usual way, giving the usual ES separation of the translational, rotational-LAM, and vibrational motions. However, additional coupling terms can be removed or minimized with judiciously chosen linear combination coefficients cj . Thereby, one can improve the usual ES separation. Let us see some details of how this comes about. The classical kinetic energy is 1  dun dun · , mn 2 n=1 dt dt N

T =

(68)

where · denotes scalar product. It is shown in the Appendix that Eq. (68) can be rewritten by using Eq. (5a) as 

  N dS dR 1 dR dR · + · T = M mn (an + d n ) 2 dt dt dt dt n=1   N

 dR d  + S · mn (an + d n ) (69a) dt dt n=1 N  1 + ω·ω mn (an + d n ) · (an + d n ) 2 n=1

1 − mn (ω · (an + d n ))2 2 n=1 N

Pn,p m1/2 p dp,

(63)

p=1

where superscript 1 indicates that projection with P has been performed only once, d p = (dxp , dyp , dzp )T , and d ES,1 n ES,1 ES,1 ES,1 T = (dxn , dyn , dzn ) . Define a transformation matrix V as V = PAP,

(64)

where A is a 3N × 3N matrix which does not commute with P. For instance, one may choose A to be the projection matrix P B corresponding to an isotopologue or a different electronic state of the molecule. By applying V j-times one obtains the

+ω ·

+ω ·

(69b)



  N  d an d an − 2ω · × dn mn a n × mn dt dt n=1 n=1

N 

N  n=1

mn

d dn dn × dt



  N d  +ω · mn (an × d n ) dt n=1

(69c)

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234107-10

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

 N d an d an d 2 an 1 · − 2 2 · dn mn 2 n=1 dt dt dt   N N d  d an 1  d dn d dn · dn + · , (69d) + mn mn dt n=1 dt 2 n=1 dt dt

+

where ω is an angular velocity vector whose components are S−1 defined as elements of the skew symmetric matrix S d dt . The displacements d n must be constrained as explained in Sec. II A. We may choose displacements d ES n obeying the ES conditions. Then the terms given in square brackets in Eq. (69) vanish and T simplifies as T =

1 dR dR M · 2 dt dt

(70a)

N      1 · an + d ES + ω·ω mn an + d ES n n 2 n=1

2 1   − mn ω · an + d ES n 2 n=1

Thus, Eq. (71) can be replaced by = m−1/2 d ES,j n n

+ ω·



Its substitution into Eq. (67) gives ⎡ ⎤ 3N−6−L 3N−6−L   )⎦ −1/2 ⎣ d ES cj l (j n = mn n,r Qr .

mn a n ×

n=1

+ ω·

− 2ω ·

mn

n=1

 

d ES d mn d ES × n dt n n=1

1 + mn 2 n=1



N

N

d an × d ES n dt



(70c)

d 2 an d an d an · − 2 2 · d ES n dt dt dt





 d ES d ES · . d d dt n dt n

(70d)

This is the kinetic energy expression when the ES conditions have been imposed in the usual fashion. The derivation would follow by transforming from displacement coordinates to normal coordinates via Eq. (21). The most general form of ES displacements is, however, given in Eq. (67). We shall show that general ES displacements can be expressed as linear combinations of a single set of normal coordinates by an expression similar to Eq. (21). , There is a relationship similar to Eq. (21) for each d ES,j n that is, ES,j dαn = m−1/2 n

3N−6−L 

(j ) ) lαn,r Q(j r ,

(71)

r=1 (j )

where lαn,r are the components of the eigenvector corresponding to the rth non-zero eigenvalue of the matrix, V j −1 HV j −1 ,  (j ) 1/2 (j ) and Qr = N n=1 mn lαn,r dαn are the corresponding normal (j +k) (j ) in place of Qr coordinates. Even when substituting Qr (j ) (while retaining the coefficients lαn,r ) in Eq. (71) one obtains ES displacements, since Eqs. (13a)–(13c) are obeyed. There(j ) (j +k) , fore, there is no need to distinguish between Qr and Qr j (1) and we can take Qr = Qr ≡ Qr .

(73)

Equation (73) is very similar to Eq. (21) indeed, with the notable exception that it contains as yet unknown coefficients cj . Having substituted Eq. (73) into Eq. (70) one may fix the c coefficients by requiring some of the coupling terms to vanish or to be minimized. Particular importance is of the Coriolis term, the last term in Eq. (70c). By ignoring terms contributing to rotational-LAM velocity (momentum) coupling, it reads as TCor = ω ·

3N−6−L 

˙ p, ζ rp Qr Q

(74)

r,p=1

where ζ rp =

3N−6−L  i,j =1

N 

1 mn 2 n=1

+

d an dt

N 

(72)

j =1

r=1

(70b)



) l (j n,r Qr .

r=1

N

N 

3N−6−L 

N 

l (i) n,r

×

) l (j n,p

ci cj .

(75)

n=1

There are 3(3N − 6 − L)2 Coriolis coefficients. The “diagonal” coefficients ζ rr = 0 and there may be Coriolis coefficients which are equal to zero due to symmetry reasons. Denote the set of the remaining Coriolis coefficients by C. Split C into two sets. Let the first set, C1 , contain those coefficients which we would like to make vanish, and let the second, C2 , contain the rest of the coefficients. The coefficients in C1 may be made zero by choosing suitable ci s, that is, by finding non-trivial solutions of the corresponding system of homogeneous quadratic polynomial equations. Since symmetry requirements prescribe relations among the ci coefficients, these must be taken into account when solving the quadratic system. While symmetry will reduce the number of independent variables, the number of equations is also reduced (halved, in fact) since ζ rp = −ζ pr . Actually one should do constrained optimization: Determine a set c which minimizes the magnitude of the Coriolis coefficients belonging to C2 under the condition that the Coriolis coefficients in C1 vanish. Note that by setting c1 = 1, c2 = 0, c3 = 0, . . . , c3N − 6 − L = 0 Eq. (74) simplifies to the usual expression of TCor . In order for this set to give possibly the optimal TCor , it should solve at least one of the problems of constrained optimization which one can set up by splitting C into two sets in different ways. Unless this requirement is met, the separation of the rotationalLAM and the vibrational motions may be enhanced with respect to standard ES separation by suitable choice of the c coefficients. The removal or minimization of terms other than Coriolis terms may also be considered. One may wonder if all Coriolis coefficients could be made vanish by suitably chosen cs (at least one of the cs should be different from zero). In absence of symmetry, there are more equations than unknowns and it is unlikely to

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234107-11

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

exist such a set of cs. When the molecule is of higher symmetry there may be fewer equations than unknowns, and there might exist a suitable set c. This would mean that the Coriolis coupling vanishes over a finite range of the vibrational coordinates which contradicts to a result of the analysis of Coriolis coupling described in Ref. 10. No definite conclusion can be drawn at this stage, more detailed investigation is required. The result described in this section is an immediate consequence of introducing ES displacements via projection (Sec. II B). Another interesting consequence is pointed out below. VII. HOW TO OBTAIN A ROTATIONAL-VIBRATIONAL HAMILTONIAN GIVEN IN TERMS OF CURVILINEAR INTERNAL COORDINATES WITH OBSERVING THE ECKART CONDITIONS

Let us consider a rigid reference configuration and the instantaneous configuration of the molecule. Let the coordinates of the atoms of the reference, aβn , and the instantaneous, rβn , configuration be given in a Cartesian system of axes whose origin is in the center of mass of the reference configuration. This system will serve as the MS as becomes clear in a moment. Express the rβn as functions of internal coordinates of the molecule. Define the displacements d n as d n = r n − a0n .

(76)

Use them in Eq. (63) with L = 0 and generate an independent set of vectors according to Eq. (65). Form the Eckart displacements d En according to Eq. (67). They obey the Eckart conditions and they are given in terms of curvilinear internal coordinates. Clearly, the initially chosen system of axes plays the role of the MS, specifically that of an Eckart system. Derive the classical kinetic energy expression. Choose suitable values for the linear combination coefficients cj to improve separation of the rotational and the vibrational motions beyond that which has already been obtained with the usual application of the Eckart conditions. Write up the classical Hamiltonian by transforming from velocities to momenta in the expression of kinetic energy. Quantize to obtain the Hamiltonian operator. Apart from the presence and treatment of the c coefficients the derivation is standard whose details can be learnt, for instance, from Refs. 7, 8, and 41. The method proposed has advantages: (a) It is simple and clear how satisfaction of the Eckart conditions is ensured. For comparison, consider a recent work21 which, in essence, employs the method discussed in Sec. II A to determine vibrational displacements obeying the Eckart conditions. In this method, the rotational matrix connecting the LS and MS must be recalculated whenever the internal coordinates change their values, whereas the projection matrix generating Eckart displacements is independent of the instantaneous values of the LS and internal coordinates. (b) It does not require any special choice of internal coordinates to have the E(S) conditions satisfied. (c) It applies to molecules with arbitrary nonlinear structure. (d) Its development may be carried out with elementary mathematical tools, i.e., it does not require knowledge of geometric algebra advocated in Ref. 22 and also employed in Ref. 23.

All in all, another example of the usefulness of constructing ES displacements via the projection described in Sec. II B has been pointed out.

VIII. SUMMARY

Distinguishing overall rotation and vibrations when describing nuclear motions in a molecule is necessary whether one aims at analysing or predicting the rotational-vibrational spectrum. It appears that the best way to achieve such a separation is through the use of the Eckart or the Eckart−Sayvetz conditions. The Eckart−Sayvetz conditions apply to nonrigid molecules with assuming effective or least squares LAM coordinates. The Eckart conditions are useful for semirigid molecules, for non-rigid molecules when real LAM coordinates are employed, and when all vibrational degrees of freedom are described as if they were LAMs. The basic idea put forth in this paper is the construction of vibrational displacements satisfying the Eckart-Sayvetz conditions via projection. The ability to construct Eckart-Sayvetz displacements as projections of unconstrained displacements has led to a number of interesting results: (i) A rigorous derivation of the transformation connecting the normal coordinates corresponding to two electronic states or two isotopologues of a molecule. The transformation is simpler than those derived earlier. (ii) Based on physical arguments the axis switching transformation has been reconsidered. It led to a simplification in calculating the transformation matrix. (iii) It has been shown that the usual method of imposing the Eckart-Sayvetz conditions may give a suboptimal separation of the rotational-LAM and the vibrational motions. A method of enhancing the separation beyond that given by the standard use of the Eckart-Sayvetz conditions has been introduced. (iv) It has been sketched how one can derive a rotational-vibrational Hamiltonian in terms of curvilinear internal coordinates with observing the Eckart conditions. Among the merits of this procedure are its clarity and simplicity in ensuring satisfaction of the Eckart conditions and the possibility for enhancing the separation of the rotational and the vibrational motions beyond that achievable with the usual application of the Eckart conditions.

ACKNOWLEDGMENTS

The author is indebted to Attila Császár for carefully reading the paper.

APPENDIX: DERIVATION OF EQ. (69) n n Let us introduce tn = du · du . Then 2T = dt dt Recall the definition, Eq. (2), of un . Thus,

N n=1

m n tn .



d R d S−1 −1 d r n + rn + S tn = dt dt dt

d R d S−1 −1 d r n + rn + S · . dt dt dt

(A1)

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234107-12

Viktor Szalay

J. Chem. Phys. 140, 234107 (2014)

Since S is a rotational matrix S−1 = ST and d ST d S−1 = = dt dt



dS dt

Therefore,

T (A2)

.

tn =

× ω · ω(an + d n ) · (an + d n ) − (ω · (an + d n ))2

 

d an d an × dn + 2ω · an × − 4ω · dt dt

 d(an × d n ) d dn + 2ω · d n × + 2ω · dt dt

 2 d d an d an d an d an + · − 2 2 · dn + 2 · dn dt dt dt dt dt

Then, by employing these equations along with the properties of the scalar product one can show that

T

T dS dR dR dS d rn d rn · +S · tn = rn · S rn + dt dt dt dt dt dt

T d R d rn dS dR d rn dS r n. +2 · r n + 2S · +2 ·S dt dt dt dt dt dt

+

(A3)

dS S dt

T

  T dS T =− S . dt

dS dt

T

⎛

0

⎜ =⎜ ⎝ ωz −ωy

−ωz 0 ωx

ωy

⎞

⎟ −ωx ⎟ ⎠,

(A5)

0

where ωα are the components of an angular velocity vector ω. Since S( ddtS )T r n = ω × r n , one can rewrite Eq. (A3) as tn =

(A9)

1 G.

(A4)

That is, S( ddtS )T is skew symmetric and it can be given as

d dn d dn · . dt dt

When this expression is multiplied by mn , summed up with respect to n, and divided by 2 one obtains Eq. (69).

Since SST = I with I denoting the 3 × 3 unit matrix

dS dR dR dR d R d(an + d n ) · +2 ·(an + d n ) + 2S · dt dt dt dt dt dt

dR dR d rn d rn · + ω · ω(r n · r n ) − (ω · r n )2 + · dt dt dt dt

 d R d rn d rn dS dR · r n + 2S · + 2ω · r n × , +2 dt dt dt dt dt (A6)

by employing the relationships a · (b × c) = b · (c × a) and a × (b × c) = b(a · c) − c(a · b). Recall that r n = an + d n and note that  





d an d rn d an = ω · an × − 2ω · × dn ω · rn × dt dt dt 

d(an × d n ) d dn +ω· + ω · dn × , dt dt (A7) and d an d an d dn d dn d rn d rn · = · + · dt dt dt dt dt dt 

d 2 an d d an · d n − 2 2 · d n . (A8) +2 dt dt dt

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